Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int dx[] = {0, 1, -1, 0, 1, -1, -1, 1};
const int dy[] = {-1, 0, 0, 1, 1, 1, -1, -1};
const int N = 0;
const int MOD = 0;
const int INF = 1e9 + 10;
const long long int LLINF = 1e18 + 10;
vector<int> pessoas[100010][20];
vector<int> g[100010];
int h[100010];
vector<int> vive[100010];
int lca[100010][20];
vector<int> global;
void curioso(vector<int> &a, vector<int> &b) {
  global.clear();
  int i = 0;
  int j = 0;
  while (1) {
    if ((i == a.size() && j == b.size()) || global.size() == 10) break;
    if (i == a.size()) {
      global.push_back(b[j++]);
    } else if (j == b.size()) {
      global.push_back(a[i++]);
    } else {
      if (a[i] == b[j]) {
        global.push_back(a[i++]);
        j++;
      } else if (a[i] > b[j]) {
        global.push_back(b[j++]);
      } else {
        global.push_back(a[i++]);
      }
    }
  }
}
void dfs(int u, int pai) {
  if (u != pai) h[u] = h[pai] + 1;
  lca[u][0] = pai;
  curioso(vive[u], vive[pai]);
  pessoas[u][0] = global;
  for (int i = 1; i < 20; i++) {
    lca[u][i] = lca[lca[u][i - 1]][i - 1];
    curioso(pessoas[u][i - 1], pessoas[lca[u][i - 1]][i - 1]);
    pessoas[u][i] = global;
  }
  for (int i = 0; i < g[u].size(); i++) {
    if (g[u][i] == pai) continue;
    dfs(g[u][i], u);
  }
}
void solve(int u, int v) {
  if (h[u] < h[v]) {
    swap(u, v);
  }
  curioso(vive[u], vive[v]);
  vector<int> ans = global;
  int d = abs(h[u] - h[v]);
  for (int i = 19; i >= 0; i--) {
    if (d & (1 << i)) {
      curioso(ans, pessoas[u][i]);
      ans = global;
      u = lca[u][i];
    }
  }
  if (u == v) {
    return;
  }
  for (int i = 19; i >= 0; i--) {
    if (lca[u][i] != lca[v][i]) {
      curioso(ans, pessoas[u][i]);
      ans = global;
      curioso(ans, pessoas[v][i]);
      ans = global;
      u = lca[u][i];
      v = lca[v][i];
    }
  }
  curioso(ans, vive[lca[u][0]]);
}
void printa() {
  for (int i = 0; i < 5; i++) {
    cout << "vértice " << i + 1 << endl;
    for (int j = 0; j < 3; j++) {
      for (int k = 0; k < pessoas[i][j].size(); k++) {
        cout << pessoas[i][j][k] << ", ";
      }
      cout << "  ";
    }
    cout << endl << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int u, v, n, m, q, x, a;
  cin >> n >> m >> q;
  for (int i = 0; i < n - 1; i++) {
    cin >> u >> v;
    u--;
    v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int i = 0; i < m; i++) {
    cin >> x;
    x--;
    vive[x].push_back(i + 1);
  }
  h[0] = 0;
  dfs(0, 0);
  vector<int> ids;
  int k;
  for (int i = 0; i < q; i++) {
    cin >> u >> v >> a;
    u--;
    v--;
    solve(u, v);
    ids = global;
    k = min((int)ids.size(), a);
    cout << k << " ";
    for (int j = 0; j < k; j++) {
      cout << ids[j] << " ";
    }
    cout << "\n";
  }
  return 0;
}